Problem: Solve for $x$ and $y$ using substitution. ${5x+3y = 11}$ ${x = -3y-5}$
Explanation: Since $x$ has already been solved for, substitute $-3y-5$ for $x$ in the first equation. ${5}{(-3y-5)}{+ 3y = 11}$ Simplify and solve for $y$ $-15y-25 + 3y = 11$ $-12y-25 = 11$ $-12y-25{+25} = 11{+25}$ $-12y = 36$ $\dfrac{-12y}{{-12}} = \dfrac{36}{{-12}}$ ${y = -3}$ Now that you know ${y = -3}$ , plug it back into $\thinspace {x = -3y-5}\thinspace$ to find $x$ ${x = -3}{(-3)}{ - 5}$ $x = 9 - 5$ ${x = 4}$ You can also plug ${y = -3}$ into $\thinspace {5x+3y = 11}\thinspace$ and get the same answer for $x$ : ${5x + 3}{(-3)}{= 11}$ ${x = 4}$